468 - Key to Success

[htl]

Could someone give some critical case of input?

[htl]

I set the buffer to 3000 characters and get WA. I get PE after setting the buffer to 6000 characters. Than's strange. Now I don't know how to fix the PE.

[Guest]

Hello,
First I got RTE, fixed it, now I got WA , but this time I could not fix it.
Can anyone enlighten me?
Thanks in advance.

Here's my code:
[c]Code Removed[/c]

[Ryan Pai]

One thing I noticed is that you have:

Code: Select all

for(i=0;i<26;i++)

I assume 26 is there because there are 26 letters. But this problem says that capital are distinct from lowercase, so this should be 52

[Ryan Pai]

This is where I double posted and looked goofy.

[Guest]

Hello Ryan,
Thank you for pointing out my mistake. It was a silly one. I forgot to change that 26 to 256 after noticing that the code should be case sensitive. Now I've got AC
Regards,

[abhijit]

i get an ouput-limit-exceeded on this.
isn't it always the case for each input set, the output line is as long as the 2nd line in the input set ?

[yiuyuho]

yes, that is the case, if you still care,

[yiuyuho]

This is weird, I submitted 2 versions of code, one treating uppercase and lowercase as different letters, and one treating upper and lower as same letter, but output them according to their case. Both get PE

So which way is intened by the problem setter?

[Ali Arman Tamal]

Hello !


yiuyuho wrote:

This is weird, I submitted 2 versions of code, one treating uppercase and lowercase as different letters, and one treating upper and lower as same letter, but output them according to their case. Both get PE

Problem description says:

The outputs of two consecutive cases will be separated by a blank line.

Note: You have to print a newline even after the last test case (though it is not said).

[smilitude]

what 's wrong?
I dont know...
Do you?

Code: Select all

/*
468 - key to success 
submission 1 runtime error 2 runtime error 3 WA 4 WA
coded at 5:08pm on 6th oct 05

comment: did i make it complex? i dont think so!
*/


#include <stdio.h>
#include <string.h>
#include <ctype.h>
#define MAX 10000

int main() {
	char not_encoded[MAX+1],encoded[MAX+1];
	int i,j,len;
	int tcases;
	char flag=0;
	struct{
		char ch;
		int occ;
	}lower[52],higher[52],temp;

	scanf("%d",&tcases);
	for(;tcases>0;tcases--) {
		scanf("%s",not_encoded); 
		scanf("%s",encoded);
		len=strlen(not_encoded);
		if(flag) printf("\n");

		for(i=0;i<26;i++) {
			lower[i].ch=i+'a';
			higher[i].ch=i+'a';
			higher[i].occ=0;
			lower[i].occ=0;
		}

		for(i=26;i<52;i++) {
			lower[i].ch=i+'A'-26;
			higher[i].ch=i+'A'-26;
			higher[i].occ=0;
			lower[i].occ=0;
		}
			
			
		for(i=0;i<len;i++) {
			if(not_encoded[i]>='a' && not_encoded[i]<='z')
				lower[not_encoded[i]-'a'].occ++;
			else if(isalpha(not_encoded[i])) 
				lower[not_encoded[i]-'A'+26].occ++;
		}
		for(i=0;i<51;i++) 
			for(j=i+1;j<52;j++) 
				if(lower[i].occ<lower[j].occ) {
					temp=lower[i];
					lower[i]=lower[j];
					lower[j]=temp;
				}
		
		len=strlen(encoded);
		for(i=0;i<len;i++) {
			if(encoded[i]>='a' && encoded[i]<='z')
				higher[encoded[i]-'a'].occ++;
			else if(isalpha(not_encoded[i]))
				higher[encoded[i]-'A'+26].occ++;
		}
		for(i=0;i<51;i++) 
			for(j=i+1;j<52;j++) 
				if(higher[i].occ<higher[j].occ) {
					temp=higher[i];
					higher[i]=higher[j];
					higher[j]=temp;
				}

		for(i=0;;i++) {
			if(higher[i].occ==0) break;
			for(j=0;j<len;j++) {
				if(higher[i].ch==encoded[j]) 
					encoded[j]=lower[i].ch;
			}
		}

		puts(encoded);
		flag=1;
	}
return 0;
}
		
	

[ayon]

I am not sure about the line !!!
=> lower.ch=i+'A-26';

what is meant by 'A-26' ?

moreover this is a multiple input problem. you have to count how many testcases...

[smilitude]

Oh, thanks a lot,
thats a good bug! [ i mean thats a serious bug! you know, bug cannot be good!]

thanks...
but whats wrong with multiple input output ...?
can you give me hint??

[ayon]

The input begins with a single positive integer on a line by itself indicating the number of the cases following. say the integer is 3. then there must be 3 sets of inputs.

Code: Select all

3
                    <- empty line
abacxbacac
qqqqqrrrrssstt
                    <- empty line
abacxbacac
qqqqqrrrrssstt
                    <- empty line
abacxbacac
qqqqqrrrrssstt
EOF

Every sets must not be same, i write the same inputs only for example.

then the output must be:

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aaaaaccccbbbxx
                          <- empty line
aaaaaccccbbbxx
                          <- empty line
aaaaaccccbbbxx

i write my code as:

Code: Select all

	scanf("%d", &test);
	for(count = 0; count < test; ++count)
	{
		scanf(" %[^\n]", s);
		scanf(" %[^\n]", p);
      .........
      .........
   }

[smilitude]

i changed my code, as you told. still i got wa...